But it should be x= +- e^4
Why is it +- (im not taking the sq rt of 8?)

You're not getting a negative because of your second step where ln(x) cannot take upon negative x values whereas in the first one the x in ln(x2) can indeed take upon both negative and positive ones. So technically speaking you can go ahead with your method as long as you remember to include the opposite sign of your answer.

So i have this differentiation question,
Show that
I tried to use product rule, but the u and v are the same and i dont know how to go about it,
I tried the chain rule, but still have no clue, as there are x terms on both of them.
So basically i dont know where to get started

(Original post by asinghj)
So i have this differentiation question,
Show that
I tried to use product rule, but the u and v are the same and i dont know how to go about it,
I tried the chain rule, but still have no clue, as there are x terms on both of them.
So basically i dont know where to get started

For part C i resolved the forces to make acceleration the subject and integrated it to get velocity, and I got D as the answer for that, not sure how to work out part D, do I have to resolve again and include kv as drag?

(Original post by B_9710)
If you exponentiate both sides to begin with you get .

the question is ln x^2 = 8
im confused how you got from that to x^2 = e^8

(Original post by RDKGames)
You're not getting a negative because of your second step where ln(x) cannot take upon negative x values whereas in the first one the x in ln(x2) can indeed take upon both negative and positive ones. So technically speaking you can go ahead with your method as long as you remember to include the opposite sign of your answer.

Hi gonna jump on this thread even though it's a bit late
I would like to be a helper and a learner lol as I have done C3, C4 and M2 and am now on a gap yahh doing FP1, FP2, S1 and S2. Did pretty good in both C3 and C4 so should be able to be quite helpful there, though maybe not as helpful as people who are really good at maths (i.e. those doing it at university)

For part C i resolved the forces to make acceleration the subject and integrated it to get velocity, and I got D as the answer for that, not sure how to work out part D, do I have to resolve again and include kv as drag?

or do I do kv=ma, use the acceleration from previous section?

now I am a future biologist, so maybe not a leading authority on engineering - but I do think I've worked it out.
It makes sense the block will reach terminal velocity, right, when the forces acting forward = those acting backward.
So you set up the diagram, mg sin(a) down the slope and uN + kV up the slope, as we are adding in the air resistance term (kV). N = mg cos (a) , so you end up with this; mg sin(a) = u mg cos(a) + kV
and you rearrange it get V out, because that V is gonna be the terminal velocity. At lesser values of V, the forces down slope would be greater and the object would accelerate. Do you agree with that?? Hope you can understand it lol (I had to use a for alpha and u for mu)

Last edited by k.russell; 15-09-2016 at 23:48.
Reason: made it clearer

Not sure how else to explain it to you. You should know that you cannot have a logarithm of a negative number, therefore cannot have x as negatives. When it comes to however, then you CAN have negative x values because negatives/positives squared just become positive.

When you go from to you are essentially getting rid off all the negative x-values hence getting rid off one of the solutions. Get it?